TITLE: Maximum value of $\sin A+\sin B+\sin C$?
QUESTION [10 upvotes]: What is the maximum value of $\sin A+\sin B+\sin C$ in a triangle $ABC$. My book says its $3\sqrt3/2$ but I have no idea how to prove it. 
I can see that if $A=B=C=\frac\pi3$ then I get $\sin A+\sin B+\sin C=\frac{3\sqrt3}2$. And also according to WolframAlpha maximum is attained for $a=b=c$. But this does not give me any idea for the proof.
Can anyone help?

REPLY [4 votes]: The function $f(\alpha,\beta,\gamma):=\sin\alpha+\sin\beta+\sin\gamma$ assumes a maximum on the simplex $$S:=\bigl\{(\alpha,\beta,\gamma)\>\bigm|\>\alpha\geq0, \ \beta\geq0,\ \gamma\geq0,\ \alpha+\beta+\gamma=\pi\bigr\}\ .$$
On the other hand, if $\alpha>\beta\geq0$ one has
$$\sin\alpha+\sin\beta=2\sin{\alpha+\beta\over2}\cos{\alpha-\beta\over2}<2\sin{\alpha+\beta\over2}\ ,$$
and $\bigl({\alpha+\beta\over2},{\alpha+\beta\over2},\gamma\bigr)\in S$.
This allows to conclude that ${\rm argmax}_S f=\bigl({\pi\over3},{\pi\over3},{\pi\over3}\bigr)$, so that$$f(\alpha,\beta,\gamma)\leq {3\sqrt{3}\over2}\qquad\bigl((\alpha,\beta,\gamma)\in S\bigr)\ .$$